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(C) The focal length of a lens depends on its refractive index and the radii of curvature of its surfaces,not on the aperture size. Therefore,the foca

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$f, \frac{3I}{4}$

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(C) The focal length of a lens depends on its refractive index and the radii of curvature of its surfaces,not on the aperture size. Therefore,the focal length remains $f$.The intensity of the image formed by a lens is directly proportional to the area of the lens exposed to incident light.Intensity $I \propto 	ext{Area} (A)$$\frac{I_2}{I_1} = \frac{A_2}{A_1}$Initial area $A_1 = \pi \left(\frac{d}{2}ight)^2 = \frac{\pi d^2}{4}$.After covering the central region with a diameter $\frac{d}{2}$,the radius of the covered part is $r = \frac{d}{4}$.Area of the covered part $A_{covered} = \pi \left(\frac{d}{4}ight)^2 = \frac{\pi d^2}{16}$.Remaining exposed area $A_2 = A_1 - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.Ratio of intensities: $\frac{I_2}{I_1} = \frac{A_2}{A_1} = \frac{3\pi d^2 / 16}{\pi d^2 / 4} = \frac{3}{4}$.Thus,the new intensity $I_2 = \frac{3}{4} I$.

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I.$ If an aperture of diameter $\frac{d}{2}$ in the central region of the lens is covered by a black paper,the focal length of the lens and the intensity of the image will be,respectively:

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